An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can’s initial volume at 8 °C is 3.5 × 10-4 m3. The coefficient of volume expansion for aluminum is 69 × 10-6 (C°)-1. When the can and the liquid are heated to 80 °C, 5.8 × 10-6 m3 of liquid spills over. What is the coefficient of volume expansion of the liquid?
Vc = Vo + a*(T-To)Vo
Vc=3.5*10^-4 + 69*10^-6*(80-8)3.5*10^-4=
3.56*10^-4 m^3 = Vol. of can at 80oC.
Vl = Vc + Vs = 3.56*10^-4 + 5.8*10^-6 =
3.618*10^-4 m^3=Vol. of liquid at 80oC.
3.618*10^-4 - 3.5*10^-4 = 1.18*10^-5 m^3
= The increase in vol. of the liquid.
a*(T-To)Vo = 1.18*10^-5 m^3
a*(72)3.5*10^-4 = 1.18*10^-5 m^3
a*0.0252 = 1.18*10^-5
a = 4.68*10^-4/oC. = Coefficient of vol.
expansion of the liquid.
To solve this problem, let's break it down into steps:
Step 1: Calculate the change in volume of the aluminum can.
The change in temperature is given as 80 °C - 8 °C = 72 °C.
The coefficient of volume expansion for aluminum is given as 69 × 10-6 (C°)-1.
Using the formula: ΔV = V0 * β * ΔT, where ΔV is the change in volume, V0 is the initial volume, β is the coefficient of volume expansion, and ΔT is the change in temperature.
Substituting the given values, we have: ΔV = 3.5 × 10-4 m³ * 69 × 10-6 (C°)-1 * 72 °C.
Calculating the result, we get: ΔV ≈ 0.0019836 m³.
Step 2: Subtract the volume of the spilled liquid.
We know that 5.8 × 10-6 m³ of liquid spills over.
So, the change in volume of the aluminum can without the liquid is: ΔV_aluminum_can = ΔV - (volume of spilled liquid) = 0.0019836 m³ - 5.8 × 10-6 m³.
Step 3: Calculate the volume expansion of the liquid.
Let's assume the initial volume of the liquid in the can as V_liquid.
The change in temperature is the same for both the can and the liquid.
Using the same formula as in Step 1, we can write: ΔV_liquid = V_liquid * β_liquid * ΔT.
Where ΔV_liquid is the change in volume of the liquid, β_liquid is the coefficient of volume expansion of the liquid, and ΔT is the change in temperature (72 °C in this case).
Rearranging the equation, we have: β_liquid = ΔV_liquid / (V_liquid * ΔT).
Substituting the known values, we get: β_liquid = (5.8 × 10-6 m³) / (V_liquid * 72 °C).
Step 4: Calculate the coefficient of volume expansion of the liquid.
From Step 2, we have the value of ΔV_aluminum_can.
We also know that the initial volume of the can V_can at 8 °C is equal to the sum of the initial volume of the liquid V_liquid and ΔV_aluminum_can.
So, we can write: V_can = V_liquid + ΔV_aluminum_can.
Substituting the known values, we have: 3.5 × 10-4 m³ = V_liquid + 0.0019836 m³.
Rearranging the equation, we get: V_liquid = 3.5 × 10-4 m³ - 0.0019836 m³.
Calculating the result, we find: V_liquid ≈ -0.0016336 m³.
Now we can substitute the calculated values into the equation from Step 3 to find the coefficient of volume expansion of the liquid:
β_liquid = (5.8 × 10-6 m³) / ((-0.0016336 m³) * 72 °C).
Calculating the result, we find the coefficient of volume expansion of the liquid.
To determine the coefficient of volume expansion of the liquid, we need to calculate the change in volume of the liquid and the change in temperature.
We are given:
Initial temperature of the can and liquid (T1) = 8 °C
Final temperature of the can and liquid (T2) = 80 °C
Initial volume of the can (V1) = 3.5 × 10-4 m3
Change in volume of the liquid (ΔV) = 5.8 × 10-6 m3
Coefficient of volume expansion of aluminum (α) = 69 × 10-6 (C°)-1
First, let's calculate the change in volume of the can. The change in volume of the can can be determined using the formula:
ΔV_can = V1 * α * ΔT
Where ΔV_can is the change in volume of the can, V1 is the initial volume of the can, α is the coefficient of volume expansion of aluminum, and ΔT is the change in temperature.
ΔT = T2 - T1 = 80 °C - 8 °C = 72 °C
ΔV_can = (3.5 × 10-4 m3) * (69 × 10-6 (C°)-1) * (72 °C) = 0.01785 × 10-4 m3
Now, we subtract the change in volume of the can from the total change in volume to find the change in volume of the liquid.
ΔV_liquid = ΔV - ΔV_can = (5.8 × 10-6 m3) - (0.01785 × 10-4 m3)
ΔV_liquid = -0.018415 × 10-4 m3
The negative sign indicates that the liquid has contracted in volume.
Finally, we can calculate the coefficient of volume expansion of the liquid using the formula:
α_liquid = ΔV_liquid / (V1 * ΔT)
α_liquid = (-0.018415 × 10-4 m3) / ((3.5 × 10-4 m3) * (72 °C))
Simplifying the calculation, we find:
α_liquid ≈ -0.027 × (C°)-1
Therefore, the coefficient of volume expansion of the liquid is approximately -0.027 (C°)-1.